y=ln(1+e∧x∧2)，求dy

Question

如题所述

Answer 1

望采纳

Answer 2

请

Answer 3

g(f(x))的导数都是内导[f(x)的导数]乘上外导[g(y)的导数，其中y=f(x)]
令t=1+e^(x²)
d[ln(1+e^(x²))]=d(lnt)
=t'*(lnt)'
=(1+e^(x²))'/(1+e^(x²))
=(2xe^(x²))/(1+e^(x²))

Answer 4

Answer 5

dy=d(ln(1+e^x^2))=[1/(1+e^x^2)]d(1+e^x^2)=1/(1+e^x^2)e^x^2d(x^2)=[(2xe^x^2)/(1+e^x^2)]dx